Chapter 4: Thermonuclear Energy Source

Transcription

1 Chapter 4: Thermonuclear Energy Source Preliminaries Reaction Cross Sections and Rates Reaction Cross Sections Reaction Rates Nonresonant Reaction Rates Resonant Reaction Rates Various Reactions The p-p Reaction The p-p Chain The CNO Cycles Helium-Burning Reactions Carbon, Oxygen, and Neon Burning Silicon Burning Comments on the reactions How are heavier elements produced?

2 Outline Preliminaries Reaction Cross Sections and Rates Reaction Cross Sections Reaction Rates Nonresonant Reaction Rates Resonant Reaction Rates Various Reactions The p-p Reaction The p-p Chain The CNO Cycles Helium-Burning Reactions Carbon, Oxygen, and Neon Burning Silicon Burning Comments on the reactions How are heavier elements produced?

3 Preliminaries Most observed stars live on so-called thermonuclear fusion, which is induced by the thermal motion and forms a heavier nucleus from several lighter ones, responsible for most of the heavy elements in the universe. The energy yield per nuclear reaction, or the Q-value, results from the difference in the total masses of the particles before and after this process: Q (Σ j M j M y )c 2. The mass of a nucleus is related to its binding energy, B ε, which is defined as the minimum energy required to break up and disperse to infinity all the constituent nucleons in the nucleus. The average bounding energy per nucleon, B ε /A, is also called the binding fraction.

4 Interesting features about the binding fraction curve: With the exception of hydrogen, typical values are around 8 MeV. The short-range nuclear force of a nucleon affects only its immediate neighborhood. 62 Ni is the most tightly bound nucleus, but is not easy to produce. 56 Ni is easily produced from lighter nuclei in the alpha process in nuclear reactions in supernovae and is decayed to become 56 Fe. The increase of B ε /A with A for A < 56 is a surface effect: particles at the surface of the nucleus experience less attraction by nuclear forces than those in the interior, and the surface area increases with radius slower than the volume so that the fraction of the surface particles drops with increasing A. The dropping of the binding energy curve for A > 56 is due to the electrostatic repulsion force of protons. The zigzags are the consequence of the shell structure of the nucleus and pair effects (notice the so-called α elements).

5 Nuclear Structure A nucleus bears many resemblances to an atom: Both are many-body problems in terms of interaction between the constituent particles. Both have bound discrete energy states. In fact, the modeling of the nuclear energy states (e.g., the shell-model) resembles that for atoms. There are critical differences: No dominant mass is in a nucleus. Hence the interaction between nucleons is more like that between particles of a cloud of gas. At larger distances, the Coulomb force becomes important for charged particles. The strong force is short ranged. It is still very uncertain in its potential form. But approximately at a few cm, the force becomes strongly attractive because of the nuclear pi-meson field.

6 Comments on nuclear structure properties Quantum nuclear level or state are characterized by its angular momentum (J), parity (P), and isospin (T, or isobaric spin or charge state), representing the basic symmetries of the underlying force law that governs the binding of nucleons in a nucleus. The quantum numbers determine how an excited state will decay. When a shell is full (that is, when the nucleons have used up all of the possible sets of quantum number assignments), a nucleus of unusual stability forms. The number of protons or neutrons in such a state is called a magic number ; e.g., 2, 8, 20, 28, 50, 82, and 126. Stable Nuclides

7 Nuclear Potential For charged particles seeking either to leave or enter a nucleus, they may need to overcome the Coulomb barrier: B c = Z α Z X e 2 /R 1.44Z α Z X /R MeV, where R is the now taken to be the minimum inter-particle separation R = 1.4(A 1/3 α + A 1/3 ) fm, where fm denotes femtometer or fermi and is cm. X The Coulomb barrier above is for p-p interaction; for heavier nucleus and/or projectiles, the barrier would be larger.

8 Nuclear Potential For charged particles seeking either to leave or enter a nucleus, they may need to overcome the Coulomb barrier: B c = Z α Z X e 2 /R 1.44Z α Z X /R MeV, where R is the now taken to be the minimum inter-particle separation R = 1.4(A 1/3 α + A 1/3 ) fm, where fm denotes femtometer or fermi and is cm. X The Coulomb barrier above is for p-p interaction; for heavier nucleus and/or projectiles, the barrier would be larger. With a typical massive nuclear density, the Fermi energy is close to 25 MeV. Thus the potential depth of the typical massive nucleus is about MeV.

9 Review of Questions and Answers How does thermonuclear fusion work qualitatively?

10 Review of Questions and Answers How does thermonuclear fusion work qualitatively? Where does the energy come from?

11 Review of Questions and Answers How does thermonuclear fusion work qualitatively? Where does the energy come from? Why is a high temperature needed?

12 Review of Questions and Answers How does thermonuclear fusion work qualitatively? Where does the energy come from? Why is a high temperature needed? In which astrophysical condition does nuclear fusion occur?

13 Review of Questions and Answers How does thermonuclear fusion work qualitatively? Where does the energy come from? Why is a high temperature needed? In which astrophysical condition does nuclear fusion occur? What is nuclear fission?

14 Review of Questions and Answers How does thermonuclear fusion work qualitatively? Where does the energy come from? Why is a high temperature needed? In which astrophysical condition does nuclear fusion occur? What is nuclear fission? Why do nuclei heavier than Fe disintegrate to release energy?

15 Outline Preliminaries Reaction Cross Sections and Rates Reaction Cross Sections Reaction Rates Nonresonant Reaction Rates Resonant Reaction Rates Various Reactions The p-p Reaction The p-p Chain The CNO Cycles Helium-Burning Reactions Carbon, Oxygen, and Neon Burning Silicon Burning Comments on the reactions How are heavier elements produced?

16 Reaction Cross Sections and Rates Most thermonuclear reactions in stars proceed through an intermediate nuclear state called the compound nucleus, which is almost always in an excited state (marked with *), i.e., α + X Z Y + β or X(α, β)y where the last expression is shorthand for the net reaction, while α is called projectile (e.g., proton, neutron, or α-particle) and β is a nuclear product of similar nature.

17 Reaction Cross Sections and Rates Most thermonuclear reactions in stars proceed through an intermediate nuclear state called the compound nucleus, which is almost always in an excited state (marked with *), i.e., α + X Z Y + β or X(α, β)y where the last expression is shorthand for the net reaction, while α is called projectile (e.g., proton, neutron, or α-particle) and β is a nuclear product of similar nature. A basic assumption here is that Z forgets how it was formed and may decay or break up in any means consistent with conservation laws and selection rules: Z X + α, Y 1 + b 1, Y 2 + b 2,..., C + γ. The first is the reproduction of the entrance channel; the last indicates a decay with γ-ray emission; the others are particle decays where the b 1, b 2,... may be, e.g., neutrons, protons, α particles, etc.

18 Reaction Cross Sections and Rates Most thermonuclear reactions in stars proceed through an intermediate nuclear state called the compound nucleus, which is almost always in an excited state (marked with *), i.e., α + X Z Y + β or X(α, β)y where the last expression is shorthand for the net reaction, while α is called projectile (e.g., proton, neutron, or α-particle) and β is a nuclear product of similar nature. A basic assumption here is that Z forgets how it was formed and may decay or break up in any means consistent with conservation laws and selection rules: Z X + α, Y 1 + b 1, Y 2 + b 2,..., C + γ. The first is the reproduction of the entrance channel; the last indicates a decay with γ-ray emission; the others are particle decays where the b 1, b 2,... may be, e.g., neutrons, protons, α particles, etc. A decay with electron emission typically has negligible probability (β decay time being of order seconds or larger).

19 Uncertainty principle and Decay Rate With each of these exit modes, we can associate a mean-life for decay, τ i, and, through the uncertainty principle, a width in energy, Γ i : Γ i τ i =. Thus the more long-lived the state is, the smaller its energy width is.

20 Uncertainty principle and Decay Rate With each of these exit modes, we can associate a mean-life for decay, τ i, and, through the uncertainty principle, a width in energy, Γ i : Γ i τ i =. Thus the more long-lived the state is, the smaller its energy width is. The probability that Z will decay through the channel i is P i = 1/τ i Σ j (1/τ j ) = Γ i Γ where Γ = Σ j Γ j = Σ j (1/τ j ) is defined as the total energy width of Z.

21 Characteristics of the Decay Rates Among the possible exit channels is the reproduction of the entrance channel with its width Γ entr. The ratio Γ entr /Γ is a measure of forming the entrance particles through that channel. Γ i is proportional to the Coulomb barrier penetration factor, which can depend on the angular momentum (l), parity conservation, and other selection rules. Nuclear states of increasingly higher excitation energy are increasingly uninhibited by the Coulomb repulsion potential. Hence they break apart increasingly rapidly, and into more and more possible combinations of final particles. The larger the rate of breakup, the larger the energy width of the state accordingly to Γ = /τ. But for the lower kinetic energies in stellar interiors, the lifetimes of the breakup into charged particles are long enough for the widths to be much less than the energy separation of nuclear state in the light nuclei.

22 Reaction Cross Section Considering a particle α with a linear momentum p comes within an impact parameter distance s. The quantized angular momentum is then sp = l, where l is the angular momentum quantum number. For angular momenta between l and l = l + 1, the target area, or fractional cross section, of a ring bounded by s l and s l+1 is π( p )2 (2l + 1). The cross section can be expressed as where σ αβ = gπ π ( ) 2 (2l + 1) Γ αγ β p Γ 2 f (ε). ( ) 2 = π 2 p 2εm = ε(mev )µ barn. Here m = m α m X /(m α + m X ) = µm A is the reduced mass while µ is the reduced mass in amu and the cross section is expressed in units of barn, defined as 1 barn = cm 2.

23 The above cross section may be divided into three parts: geometric, Coulomb barrier plus nuclear physics, and the energy dependence due to the resonance.

24 The above cross section may be divided into three parts: geometric, Coulomb barrier plus nuclear physics, and the energy dependence due to the resonance. The resonant form of f (ε) is that of an isolated Breit-Wigner resonance: Γ 2 f (ε) = (ε ε r ) 2 + (Γ/2) 2 which is strongly peaked at, or near, the resonance energy ε r, related to a discrete state at an energy ε of the compound nucleus according to ε r = (m Z c 2 + ε ) (m α c 2 + m X c 2 ).

25 The above cross section may be divided into three parts: geometric, Coulomb barrier plus nuclear physics, and the energy dependence due to the resonance. The resonant form of f (ε) is that of an isolated Breit-Wigner resonance: Γ 2 f (ε) = (ε ε r ) 2 + (Γ/2) 2 which is strongly peaked at, or near, the resonance energy ε r, related to a discrete state at an energy ε of the compound nucleus according to ε r = (m Z c 2 + ε ) (m α c 2 + m X c 2 ). For energies near ε r, the f (ε) factor dominates the cross section. Therefore, when the kinetic energy of the particles at infinity is such that the total energy coincides with one of the quasi-stationary states of the compound nucleus, a resonant reaction can occur.

26 The above cross section may be divided into three parts: geometric, Coulomb barrier plus nuclear physics, and the energy dependence due to the resonance. The resonant form of f (ε) is that of an isolated Breit-Wigner resonance: Γ 2 f (ε) = (ε ε r ) 2 + (Γ/2) 2 which is strongly peaked at, or near, the resonance energy ε r, related to a discrete state at an energy ε of the compound nucleus according to ε r = (m Z c 2 + ε ) (m α c 2 + m X c 2 ). For energies near ε r, the f (ε) factor dominates the cross section. Therefore, when the kinetic energy of the particles at infinity is such that the total energy coincides with one of the quasi-stationary states of the compound nucleus, a resonant reaction can occur. The nonresonant case arises when f (ε) is slowly varying compared to other factors in the cross section, commonly arising when ε is far removed from ε r, if present.

27 Reaction Rates The total reaction rate is r αβ = n α n X < σv > αβ. (1) If α and X are the same species of particles, then the above equation should be divided by 2 to avoid double counting.

28 Reaction Rates The total reaction rate is r αβ = n α n X < σv > αβ. (1) If α and X are the same species of particles, then the above equation should be divided by 2 to avoid double counting. The above average < σv > αβ = 0 Ψ(ε)σ αβ (ε)vdε; Ψ(ε) = 2 π 1/2 (kt ) 3/2 e ε/kt ε 1/2 is taken, assuming a Maxwell-Boltzmann energy distribution in the center of mass system, which gives < σv > αβ µ1/2 (kt ) 3/2 0 σ αβ e ε/kt εdε. (2)

29 Reaction Rates The total reaction rate is r αβ = n α n X < σv > αβ. (1) If α and X are the same species of particles, then the above equation should be divided by 2 to avoid double counting. The above average < σv > αβ = 0 Ψ(ε)σ αβ (ε)vdε; Ψ(ε) = 2 π 1/2 (kt ) 3/2 e ε/kt ε 1/2 is taken, assuming a Maxwell-Boltzmann energy distribution in the center of mass system, which gives < σv > αβ µ1/2 (kt ) 3/2 The effective energy generation per mass is 0 σ αβ e ε/kt εdε. (2) ɛ = r αβ Q eff /ρ, (3) where Q eff is the effective yield, after subtracting the energy carried away by the neutrinos.

30 Nonresonant Reaction Rates The probability for two particles of charges Z α and Z X and moving at a relative velocity of v to penetrate their electrostatic repulsion is proportional to the factor (assuming B C ε), Γ α e 2πη where the dimensionless Sommerfield factor for the entrance channel, for example, is η = Z αz X e 2 v = b/ε 1/2, = Z α Z X ( µ ε )1/2 where b = 0.99Z α Z X µ 1/2 (MeV) 1/2. η represents the ratio of the Coulomb potential and the energy uncertainty as the projectile passing the ion.

31 Nonresonant Reaction Rates The probability for two particles of charges Z α and Z X and moving at a relative velocity of v to penetrate their electrostatic repulsion is proportional to the factor (assuming B C ε), Γ α e 2πη where the dimensionless Sommerfield factor for the entrance channel, for example, is η = Z αz X e 2 v = b/ε 1/2, = Z α Z X ( µ ε )1/2 where b = 0.99Z α Z X µ 1/2 (MeV) 1/2. η represents the ratio of the Coulomb potential and the energy uncertainty as the projectile passing the ion. For ε kt 0.1 MeV, µ 1 and Z α Z X 2, one finds that 2πη 12. So Γ α is very sensitive to ε.

32 Gamow Peak The situation for the exit channel is different. Even for charged particle channels, stellar exothermic reactions with Q-values in excess of a couple of MeV mean that the Coulomb barrier is relatively easy to tunnel through. Thus Γ β can be taken to be a constant or to vary slowly with energy. The cross-section can thus be expressed as σ αβ Γ αγ β S(ε) 1/2 f (ε) = µεγ2 µε e b/ε, where S(ε) is a slowly varying function of ε.

33 Gamow Peak The situation for the exit channel is different. Even for charged particle channels, stellar exothermic reactions with Q-values in excess of a couple of MeV mean that the Coulomb barrier is relatively easy to tunnel through. Thus Γ β can be taken to be a constant or to vary slowly with energy. The cross-section can thus be expressed as σ αβ Γ αγ β S(ε) 1/2 f (ε) = µεγ2 µε e b/ε, where S(ε) is a slowly varying function of ε. Averaging over the Maxwell-Boltzmann energy distribution, we have < σv > αβ 1 µ 1/2 (kt ) 3/2 The integrand is aptly called the Gamow Peak. 0 S(ε)e ε/kt b/ε1/2 dε.

34 Gamow Peak To see this clearly, we consider the cross section in a general form S(ε)e f (ε) dε where the function f (ε) has a value much larger than unity and a single sharp minimum at ε 0. The integral may be approximated by expanding f (ε) f (ε) f (ε 0 ) + f (ε 0 ) (ε ε 0) where f (ε 0 ) = 0 has been assumed. Then a good estimate for the value of the integral becomes S(ε 0 )e f (ε 0) exp [ f (ε 0 ) (ε ε 0) 2 2 ] dε = 2π f (ε 0 ) S(ε 0)e f (ε 0).

35 In our case, ε 0 = (bkt /2) 2/3 = 1.22(ZαZ 2 X 2µT 6 2)1/3 kev ( ) f (ε 0 ) = ε 0 /kt + b/ε 1/2 Z 2 α 0 = 42 Z 2 1/3 X µ T 6 f 3 (ε 0 ) =. 2 1/3 b 2/3 (kt ) 5/3 The approximate full width of the Gamow peak (at 1/e of the maximum) is 23/2 f (ε 0 ) = 0.75(Z 2 αz 2 X µt 5 6 )1/6 kev.

36 In our case, ε 0 = (bkt /2) 2/3 = 1.22(ZαZ 2 X 2µT 6 2)1/3 kev ( ) f (ε 0 ) = ε 0 /kt + b/ε 1/2 Z 2 α 0 = 42 Z 2 1/3 X µ T 6 f 3 (ε 0 ) =. 2 1/3 b 2/3 (kt ) 5/3 The approximate full width of the Gamow peak (at 1/e of the maximum) is 23/2 f (ε 0 ) = 0.75(Z 2 αz 2 X µt 5 6 )1/6 kev. Finally, we have ( < σv > αβ = ( cm 3 s 1 Zα Z ) X 1/3e Z 2 42( α Z 2 ) 1/3 X µ T )S(ε 0 ) 6 µt6 2.

37 In our case, ε 0 = (bkt /2) 2/3 = 1.22(ZαZ 2 X 2µT 6 2)1/3 kev ( ) f (ε 0 ) = ε 0 /kt + b/ε 1/2 Z 2 α 0 = 42 Z 2 1/3 X µ T 6 f 3 (ε 0 ) =. 2 1/3 b 2/3 (kt ) 5/3 The approximate full width of the Gamow peak (at 1/e of the maximum) is 23/2 f (ε 0 ) = 0.75(Z 2 αz 2 X µt 5 6 )1/6 kev. Finally, we have ( < σv > αβ = ( cm 3 s 1 Zα Z X )S(ε 0 ) µt 2 6 ) 1/3e 42( Z 2 α Z 2 X µ T 6 ) 1/3. The energy of individual thermal particles in a star is low, compared to the energy range used in laboratory determinations of the reaction cross sections. Certain extrapolation of the experiment-determined cross-section to low energies is often needed.

38 The common procedure is to extract σ αβ at experimentally accessible energies (usually much below 100 kev) and plot S(ε) = σ αβ εe 2πη. If necessary, S(ε) is then extrapolated, with perhaps some help from theory, down to astrophysical interesting energies. S(ε) is only weakly dependent of energy, changing by less than 10% of the cross section change between 0 and 100 kev.

39 The common procedure is to extract σ αβ at experimentally accessible energies (usually much below 100 kev) and plot S(ε) = σ αβ εe 2πη. If necessary, S(ε) is then extrapolated, with perhaps some help from theory, down to astrophysical interesting energies. S(ε) is only weakly dependent of energy, changing by less than 10% of the cross section change between 0 and 100 kev. Notice that for the non-resonant reaction, ɛ ρ T 2/3 6 1/3 b/t e 6, To use the expression ɛ = ɛ 0 ρ λ T ν, one obtain λ = 1 and ν = (for the p-p reaction and T 6 15 as in the Sun). b 3T 1/3 6

40 The common procedure is to extract σ αβ at experimentally accessible energies (usually much below 100 kev) and plot S(ε) = σ αβ εe 2πη. If necessary, S(ε) is then extrapolated, with perhaps some help from theory, down to astrophysical interesting energies. S(ε) is only weakly dependent of energy, changing by less than 10% of the cross section change between 0 and 100 kev. Notice that for the non-resonant reaction, ɛ ρ T 2/3 6 1/3 b/t e 6, To use the expression ɛ = ɛ 0 ρ λ T ν, one obtain λ = 1 and ν = (for the p-p reaction and T 6 15 as in the Sun). b 3T 1/3 6 The nonresonant reaction rate increases with the increasing temperature and deceases with the increasing charge of the interacting particles. But this monotonic trend can be interfered with the resonant reactions.

41 Resonant Reaction Rates Resonant reactions occur when the energy of the interacting particles corresponds to an energy level of the compound nucleus. In this case, the reaction cross section can be orders of magnitude higher than those at neighboring energies and is dominated by the factor Γ 2 f (ε) = (ε ε r ) 2 + (Γ/2) 2. Other factors, Ψ(ε) and Γs, vary slowly over a resonance. Therefore, < σv > αβ (kt ) 3/2 e εr /kt dε Γ α (ε r )Γ β (ε r ) 0 (ε ε r ) 2 + (Γ/2) 2. It is customary to extend the lower limit of the integrand to. Then < σv > αβ (kt ) 3/2 εr /kt e where the coefficient, proportional to the total cross-section of the resonance, is often determined experimentally. Similar to the 11.6εr (MeV) non-resonant reaction, we can get the exponent ν = T

42 Outline Preliminaries Reaction Cross Sections and Rates Reaction Cross Sections Reaction Rates Nonresonant Reaction Rates Resonant Reaction Rates Various Reactions The p-p Reaction The p-p Chain The CNO Cycles Helium-Burning Reactions Carbon, Oxygen, and Neon Burning Silicon Burning Comments on the reactions How are heavier elements produced?

43 The p-p Reaction One of the prime example of the nonresonant reaction is p + p 2 D + e + + ν e. This crucial p-p reaction requires that two protons form a coupled system (the diproton ) while flashing past one another: one of them must undergo a weak decay by emitting a positron and electron neutrino (there is no bound nucleus 2 He). The two remaining particles, proton and neutron, are then left together as the rather fragile Deuterium (with a binding energy of 2.22 MeV).

44 The p-p Reaction One of the prime example of the nonresonant reaction is p + p 2 D + e + + ν e. This crucial p-p reaction requires that two protons form a coupled system (the diproton ) while flashing past one another: one of them must undergo a weak decay by emitting a positron and electron neutrino (there is no bound nucleus 2 He). The two remaining particles, proton and neutron, are then left together as the rather fragile Deuterium (with a binding energy of 2.22 MeV). The reaction rate is r pp = T 2/3 9 X 2 ρ 2 1/ /T e 9 cm 3 s 1 (4)

45 The p-p Reaction One of the prime example of the nonresonant reaction is p + p 2 D + e + + ν e. This crucial p-p reaction requires that two protons form a coupled system (the diproton ) while flashing past one another: one of them must undergo a weak decay by emitting a positron and electron neutrino (there is no bound nucleus 2 He). The two remaining particles, proton and neutron, are then left together as the rather fragile Deuterium (with a binding energy of 2.22 MeV). The reaction rate is r pp = T 2/3 9 X 2 ρ 2 1/ /T e 9 cm 3 s 1 (4) The mean life of a proton against the p-p reaction destruction is τ p = n p dn p /dt = n p. 2r pp For T , ρ 100 g cm 3, and X 0.7, τ p yrs, close to the lifetime of the sun!

46 Try to write down the reactions that end up with the production of α particles (He nuclides). Keep in mind that the most abundant projectiles at this stages are p, e, He and its isotopes.

47 The MAJOR reaction sequences, appropriate at normal MS stellar temperatures, are the following: PP I : p + p 2 D + e + + ν e 2 D + p 3 He + γ 3 He + 3 He 4 He + 2p

48 The MAJOR reaction sequences, appropriate at normal MS stellar temperatures, are the following: PP I : p + p 2 D + e + + ν e 2 D + p 3 He + γ 3 He + 3 He 4 He + 2p PP II : 3 He + 4 He 7 Be + γ 7 Be + e 7 Li + ν e + γ 7 Li + p 2 4 He

49 The MAJOR reaction sequences, appropriate at normal MS stellar temperatures, are the following: PP I : p + p 2 D + e + + ν e 2 D + p 3 He + γ 3 He + 3 He 4 He + 2p PP II : 3 He + 4 He 7 Be + γ 7 Be + e 7 Li + ν e + γ 7 Li + p 2 4 He PP III : 7 Be + p 8 B + γ 8 B 8 Be + e + + ν e 8 Be 2 4 He

50 The MAJOR reaction sequences, appropriate at normal MS stellar temperatures, are the following: PP I : p + p 2 D + e + + ν e 2 D + p 3 He + γ 3 He + 3 He 4 He + 2p PP II : 3 He + 4 He 7 Be + γ 7 Be + e 7 Li + ν e + γ 7 Li + p 2 4 He PP III : 7 Be + p 8 B + γ 8 B 8 Be + e + + ν e 8 Be 2 4 He These chains become more important in the order I, II, and III as temperature increases (e.g., 7 Be + p 8 B + γ requires to overcome a larger B c than 7 Be + e 7 Li + ν e + γ).

51 Energy generation via the p-p Chain Eventually, four protons are used to make each α particle, and two of the protons must be converted to neutrons (via weak reactions). The p-p-reaction is the slowest, by a substantial margin, and hence controls the rate of the chains. Therefore, other reactions are all kind of in equilibrium states. The energy generation rate for the p-p chain must be proportional to < σv > of the p-p reaction. The overall effective Q-value for the chain depends on the weighted contributions of the three sub-chains to the rate of processing. Each of these contributes differently because of the quantities and energies of the neutrinos lost. The overall effective Q-value, 13 MeV, of the p-p chain in conjunction with Eqs. 3 and 4, gives ɛ pp = T 2/3 9 X 2 1/ /T ρe 9 erg g 1 s 1

52 Periodic table

53 The CNO Cycles The major reactions comprising the CNO cycles at normally occurring hydrogen-burning temperatures are: 12 C + p 13 N + γ 13 N 13 C + e + + ν e 13 C + p 14 N + γ 14 N + p 15 O + γ 15 O 15 N + e + + ν e 15 N + p 12 C + 4 He 13 C + p 10 B + He is not considered because the B c is too small for B.

54 The CNO Cycles The major reactions comprising the CNO cycles at normally occurring hydrogen-burning temperatures are: 12 C + p 13 N + γ 13 N 13 C + e + + ν e 13 C + p 14 N + γ 14 N + p 15 O + γ 15 O 15 N + e + + ν e 15 N + p 12 C + 4 He 15 N + p 16 O + γ 16 O + p 17 F + γ 13 C + p 10 B + He is not considered because the B c is too small for B. 17 F 17 O + e + + ν e 17 O + p 14 N + 4 He

55 The CNO Cycles The major reactions comprising the CNO cycles at normally occurring hydrogen-burning temperatures are: 12 C + p 13 N + γ 13 N 13 C + e + + ν e 13 C + p 14 N + γ 14 N + p 15 O + γ 15 O 15 N + e + + ν e 15 N + p 12 C + 4 He 15 N + p 16 O + γ 16 O + p 17 F + γ 13 C + p 10 B + He is not considered because the B c is too small for B. 17 F 17 O + e + + ν e 17 O + p 14 N + 4 He The first set is called CN cycle, while the first and second combined is the CNO cycle.

56 The branching to this third segment is somewhat uncertain because the rate for the reaction 17 O(p, α) 14 N is not well determined. 17 O + p 18 F + γ 18 F + e 18 O + ν e 18 O + p 19 F + γ 19 F + p 16 O + 4 He

57 The branching to this third segment is somewhat uncertain because the rate for the reaction 17 O(p, α) 14 N is not well determined. 17 O + p 18 F + γ 18 F + e 18 O + ν e 18 O + p 19 F + γ 19 F + p 16 O + 4 He The above sets of reactions can be summarized as 12 C + 4p 12 C + 4 He (including 14 N + 2p 12 C + 4 He)

58 The branching to this third segment is somewhat uncertain because the rate for the reaction 17 O(p, α) 14 N is not well determined. 17 O + p 18 F + γ 18 F + e 18 O + ν e 18 O + p 19 F + γ 19 F + p 16 O + 4 He The above sets of reactions can be summarized as 12 C + 4p 12 C + 4 He (including 14 N + 2p 12 C + 4 He) or 16 O, which +2p 14 N + 4 He

59 The branching to this third segment is somewhat uncertain because the rate for the reaction 17 O(p, α) 14 N is not well determined. 17 O + p 18 F + γ 18 F + e 18 O + ν e 18 O + p 19 F + γ 19 F + p 16 O + 4 He The above sets of reactions can be summarized as 12 C + 4p 12 C + 4 He (including 14 N + 2p 12 C + 4 He) or 16 O, which +2p 14 N + 4 He or 18 F, which +2p 16 O + 4 He.

60 The branching to this third segment is somewhat uncertain because the rate for the reaction 17 O(p, α) 14 N is not well determined. 17 O + p 18 F + γ 18 F + e 18 O + ν e 18 O + p 19 F + γ 19 F + p 16 O + 4 He The above sets of reactions can be summarized as 12 C + 4p 12 C + 4 He (including 14 N + 2p 12 C + 4 He) or 16 O, which +2p 14 N + 4 He or 18 F, which +2p 16 O + 4 He. The isotopes of the CNO elements are catalysts, in the sense that they are destructed and created within the cycles. However, the concentrations do change with time, depending primarily on the relative rates of the reactions as well as the initial abundances of the elements.

61 Comments on the CNO Cycle A key reaction in the cycles is 14 N(p, γ) 15 O, which is relatively slow (hence setting the pace for the whole) and involves 14 N that appears in the first two cycles. As a result, the most abundant metal nuclides in a star that undergoes the CNO cycles will end up being 14 N. Virtually all 14 N in nature is produced in this way. In general, the C, N, and O account for the bulk of the metals in terms of the mass. Often the stellar evolution is fast enough that the CN or CNO cycles cannot reach equilibrium, resulting time-dependent isotope ratios. Many highly evolved stars show abundance ratios between 12 C, 13 C, and 14 N in their spectra that are anomalous, compared to some sort of cosmic standard.

62 CNO Energy Generation Rate The energy generation rate for the CN or CNO cycles is ɛ CNO = ρxz 1/ /T e T 2/3 9 erg g 1 s 1 9 The temperature exponent is ν(cno) = /3 T 1/3 6 which is about 18 for T 6 = 20.

63 CNO Energy Generation Rate The energy generation rate for the CN or CNO cycles is ɛ CNO = ρxz 1/ /T e T 2/3 9 erg g 1 s 1 9 The temperature exponent is ν(cno) = /3 T 1/3 6 which is about 18 for T 6 = 20. For a solar central temperature of T 6 15, X = 0.7, and Z = 0.02, ɛ pp 10ɛ CNO. But this dominance of the pp-chain in the energy generation disappears at a higher temperature because of the greater temperature sensitivity of the CNO cycle.

64 Helium-Burning Reactions: the triple α reaction The primary process in Helium-burning is the triple α reaction: 2α 8 Be(α, γ) 12 C The reaction 2α 8 Be is the inverse of 8 Be 2α that terminates the PP-III chain and is thus endothermic (energy absorbing; Q = 92 kev). This resonance can be encroached upon by the Gamow peak when the temperature reaches K. The rate of the 2α 8 Be reaction becomes high enough to make up for the short lifetime of 8 Be (10 16 s), approaching the equilibrium, The level diagram and energetics of the two reactions composing the triple-α reaction (not to scale). n 2 α < σv > αα /2 n( 8 Be)Γ decay / or 2α 8 Be.

65 Nuclear Saha equation Similar to the atomic ionization reactions, we can figure out the concentration by using the Saha equation, by making the replacements: (n +, n e ) n α and n 0 n( 8 Be), g = 1 for both α and the ground state of 8 Be, χ H Q, and the reduced mass m 2 α/m( 8 Be) m α /2. nα 2 ( ) 3/2 n( 8 Be) = πmα kt h 2 e Q/kT = T 3/2 9 e 1.065/T 9 For typical conditions at, say, the start of the helium flash in lower mass stars where ρ 10 6 g cm 3 (n α cm 3 ) and T 9 0.1, n(8 Be) n α 10 8.

66 Second stage of the triple α reaction With this seed of 8 Be nuclei, the second stage of the triple α reaction, 8 Be(α, γ) 12 C, may now continue, which is a resonant reaction with Q = MeV [= (m Be + m α M12 C)c 2 ], via an excited state 12 C with zero spin at MeV (above M12 Cc 2 ). Thus the required kinetic energy is 0.29 MeV. The emission of a γ ray photon by 12 C, however, does not come easily because it almost always decays right back to 8 Be. But the forward reaction is sufficient fast so that a small pool of 12 C is built up, according to the corresponding nuclear Saha equation, which can be used to get n( 12 C ). The net rate of decay of 12 C by γ ray cascade is n( 12 C )Γ rad /, where Γ rad = 3.67 MeV.

67 Energy generation of the triple-α reaction The overall rate of the reaction is controlled by the formation rate of the ground state of 12 C. The total energy yield is Q = =7.275 MeV and the energy generation rate is ɛ 3α = Q (ρy /(4m A)) 3 3!ρ < σv > 3α = ρ 2 Y 3 T 3 e 4.4/T 9 erg g 1 s 1, where the factor 4.4 comes from ε r = = 0.38 MeV for the formation of a 12 C from a triple-α reaction (actually two resonance reactions), A α = 4, and 3! accounts for

68 Energy generation of the triple-α reaction The overall rate of the reaction is controlled by the formation rate of the ground state of 12 C. The total energy yield is Q = =7.275 MeV and the energy generation rate is ɛ 3α = Q (ρy /(4m A)) 3 3!ρ < σv > 3α = ρ 2 Y 3 T 3 e 4.4/T 9 erg g 1 s 1, where the factor 4.4 comes from ε r = = 0.38 MeV for the formation of a 12 C from a triple-α reaction (actually two resonance reactions), A α = 4, and 3! accounts for the triple counting among the three indistinguishable α-particles. The temperature and density exponents are λ 3α = 2 and ν 3α = 4.4/T for T 9 = 0.1, which is considerably larger than the corresponding exponent for hydrogen-burning, hence potentially more explosive than hydrogen.

69 Other Helium(α)-burning reactions 12 C(α, γ) 16 O is more difficult to deal with. The 12 C + α pair, at zero initial energy, enters 16 O at 7.12 MeV (Q-value), whereas the nearest resonance in 16 O lies some 45 kev below. Hence the reaction proceeds only in the upper tail of the resonance.

70 Other Helium(α)-burning reactions 12 C(α, γ) 16 O is more difficult to deal with. The 12 C + α pair, at zero initial energy, enters 16 O at 7.12 MeV (Q-value), whereas the nearest resonance in 16 O lies some 45 kev below. Hence the reaction proceeds only in the upper tail of the resonance. The nuclear parameters for this resonance is hard to come by experimentally. There are (at least) two levels well above the entry point that can contribute. Thus only a theoretical calculated cross section can be used with quite some uncertainty. The competition between how fast 12 C is produced and how quickly it is converted to 16 O primarily determines their final relative abundances. This may also determine whether the final core, as a white dwarf, is mostly carbon or oxygen. The amounts of 16 O made also control the production of heavier elements in later burning stages.

71 Other Helium(α)-burning reactions 12 C(α, γ) 16 O is more difficult to deal with. The 12 C + α pair, at zero initial energy, enters 16 O at 7.12 MeV (Q-value), whereas the nearest resonance in 16 O lies some 45 kev below. Hence the reaction proceeds only in the upper tail of the resonance. The nuclear parameters for this resonance is hard to come by experimentally. There are (at least) two levels well above the entry point that can contribute. Thus only a theoretical calculated cross section can be used with quite some uncertainty. The competition between how fast 12 C is produced and how quickly it is converted to 16 O primarily determines their final relative abundances. This may also determine whether the final core, as a white dwarf, is mostly carbon or oxygen. The amounts of 16 O made also control the production of heavier elements in later burning stages. 16 O(α, γ) 20 Ne is rather slow because no appropriate resonance in 20 Ne is available nearby where the 16 O + α pair enters 20 Ne.

72 Carbon, Oxygen, and Neon Burning The helium-burning uses up α particles and generate Carbon, Oxygen, plus some Neon. If temperatures can rise to T 9 = 0.5 1, then carbon-burning commences and, at T 9 1, oxygen burning. The important branches of these reactions are Reaction Yield Q(MeV) 12 C + 12 C 20 Ne + α 44% C + 12 C 23 Na + p 56% O + 16 O 28 Si + α 21% O + 16 O 31 P + p 61% O + 16 O 31 S + n 18% 1.50 These reactions release p, n, and α-particles as well heavier nuclei. Such projectile particles can also be made free photo-disintegration, via photons from the tail of the Planck distribution. The photo-disintegration breaks down Ne just produced, via 20 Ne(γ, α) 16 O, which is the inverse of the last reaction in Helium-burning.

73 The generated particles can then be relatively easily fuse with available nuclides to produce even heavier ones. Examples are 23 Na(p, γ) 24 Mg during carbon burning and 31 S 31 P + e + + ν e (Phosphorus), 31 P(p, α) 28 Si(α, γ) 32 S, depending on temperature and density, even yielding 42 Ca and 46 Ti (Titanium ). Furthermore, the α-particle produced can be captured right back by 20 Ne, allowing the sequence 20 Ne(α, γ) 24 Mg(α, γ) 28 Si. The net result of these results is a pool of 16 O, 24 Mg, and 28 Si.

74 Silicon Burning When T 9 3, a bewildering number of reactions are possible. The photo-disintegration, in particular, can lead such reactions as 28 Si(γ, α) 24 Mg, followed by 24 Mg(α, p) 27 Al(α, p) 30 Si, eventually lead up to nuclei in the iron peak. The reactions are sufficient rapidly that a state of quasi-static equilibrium can be reached. Thus, the nuclear Saha equation can be used to estimate the concentrations of individual elements. The network of reactions is called Silicon burning (or perhaps melting ). Above 40 20Ca, the Z A/2 no longer applies for stable nuclides, because of the increasing Coulomb repulsive force. Heavier nuclides tend to be more neutron-rich. If enough time is allow to elapse (as in quiescent burning), then the most abundant of the elements made is 56 26Fe (not an α element). If, on the other hand, the time scales are short as in a supernova and electron and positron decays and electron captures (weak reactions) do not have enough time to go to completion, then 56 Ni (an α element) is the most abundant.

75 Characteristic Temperatures and Exponents of Various Reactions Reaction T (MeV) ν p-p CNO α C,Ne,O-burning ( 16 O + 16 O) Si-burning 3

76 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements.

77 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements. For examples, two 12 C do not often make 24 Mg and instead to form 20 Ne first. Two 28 Si can hardly fuse, because the barrier is too high.

78 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements. For examples, two 12 C do not often make 24 Mg and instead to form 20 Ne first. Two 28 Si can hardly fuse, because the barrier is too high. The fusion is mostly mediated by α particles, forming α nuclei, having (in effect) many α particles stuck together: 12 C, 16 O, 20 Ne, 24 Mg, 28 Si, 32 S, and 56 Ni.

79 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements. For examples, two 12 C do not often make 24 Mg and instead to form 20 Ne first. Two 28 Si can hardly fuse, because the barrier is too high. The fusion is mostly mediated by α particles, forming α nuclei, having (in effect) many α particles stuck together: 12 C, 16 O, 20 Ne, 24 Mg, 28 Si, 32 S, and 56 Ni. The stable 40 Ar and 56 Fe are not α nuclei. The latter is the β-decay product of 56 Ni.

80 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements. For examples, two 12 C do not often make 24 Mg and instead to form 20 Ne first. Two 28 Si can hardly fuse, because the barrier is too high. The fusion is mostly mediated by α particles, forming α nuclei, having (in effect) many α particles stuck together: 12 C, 16 O, 20 Ne, 24 Mg, 28 Si, 32 S, and 56 Ni. The stable 40 Ar and 56 Fe are not α nuclei. The latter is the β-decay product of 56 Ni. The capturing of free protons and neutrons form smaller amount adjacent nuclides, reaching kind of nuclear statistical equilibrium, or the e-process.

81 Comments on the reactions The path of nucleosynthesis depends on the right energy level, spin, parity and other roles of nuclear reactions as well as the stability of a synthesized element, the Coulomb barrier, and abundances of participating elements. For examples, two 12 C do not often make 24 Mg and instead to form 20 Ne first. Two 28 Si can hardly fuse, because the barrier is too high. The fusion is mostly mediated by α particles, forming α nuclei, having (in effect) many α particles stuck together: 12 C, 16 O, 20 Ne, 24 Mg, 28 Si, 32 S, and 56 Ni. The stable 40 Ar and 56 Fe are not α nuclei. The latter is the β-decay product of 56 Ni. The capturing of free protons and neutrons form smaller amount adjacent nuclides, reaching kind of nuclear statistical equilibrium, or the e-process. But processes like the Si burning, with an energy yield of 0.1 MeV per nucleon, occurs on scales of a day, compared to the half-life of seven days for 56 Ni. Thus no equilibrium is reached for 56 Ni.

82 Neutrino production In addition to various mentioned processes, there are other processes that we have not mentioned. In particular, the neutrino emission plays a fundamental role in our understanding of the stellar physics. The combined neutrino loss rates (in ergs g 1 s 1 ) versus ρ/µ e and temperature. At high temperatures, energy loss due to escaping neutrinos becomes increasingly important. For 10 9 K, the luminosity can be much greater than L! The pp-chain and neutrino production (see the story told by John Bahcall; jnb/papers/popular/s

83 Outline Preliminaries Reaction Cross Sections and Rates Reaction Cross Sections Reaction Rates Nonresonant Reaction Rates Resonant Reaction Rates Various Reactions The p-p Reaction The p-p Chain The CNO Cycles Helium-Burning Reactions Carbon, Oxygen, and Neon Burning Silicon Burning Comments on the reactions How are heavier elements produced?

84 How are heavier elements produced? slow s-process neutron capture by pre-existing iron peak nuclei (mostly 56 Fe) during the regular CNO fusion, which produce neutron-rich isotopes, while free neutrons can then be generated from such reactions as 13 C + α 16 O + n. The rate of neutron capture by nuclei is slow relative to the rate of radioactive β-decay. A stable isotope captures a neutron, and the resultant radioactive isotope decays to its stable daughter before the next neutron is captured. Nuclides (isotopes). The red vertical and horizontal lines show the magic numbers, reflecting regions where nuclei are expected to be more tightly bound and have longer half-lives.

85 How are heavier elements produced? slow s-process neutron capture by pre-existing iron peak nuclei (mostly 56 Fe) during the regular CNO fusion, which produce neutron-rich isotopes, while free neutrons can then be generated from such reactions as 13 C + α 16 O + n. The rate of neutron capture by nuclei Nuclides (isotopes). The red vertical is slow relative to the rate of and horizontal lines show the magic radioactive β-decay. numbers, reflecting regions where A stable isotope captures a neutron, nuclei are expected to be more tightly and the resultant radioactive isotope bound and have longer half-lives. decays to its stable daughter before the next neutron is captured. This process produces stable isotopes by moving along the valley of β-decay stable isobars in the chart of isotopes.

86 How are heavier elements produced? slow s-process neutron capture by pre-existing iron peak nuclei (mostly 56 Fe) during the regular CNO fusion, which produce neutron-rich isotopes, while free neutrons can then be generated from such reactions as 13 C + α 16 O + n. The rate of neutron capture by nuclei Nuclides (isotopes). The red vertical is slow relative to the rate of and horizontal lines show the magic radioactive β-decay. numbers, reflecting regions where A stable isotope captures a neutron, nuclei are expected to be more tightly and the resultant radioactive isotope bound and have longer half-lives. decays to its stable daughter before the next neutron is captured. This process produces stable isotopes by moving along the valley of β-decay stable isobars in the chart of isotopes. The s-process produces approximately half of the isotopes of the elements heavier than iron.

87 rapid r-process a nucleosynthesis process, that occurs in core-collapse supernovae and is responsible for the creation of approximately half of the neutron-rich atomic nuclei heavier than iron.

88 rapid r-process a nucleosynthesis process, that occurs in core-collapse supernovae and is responsible for the creation of approximately half of the neutron-rich atomic nuclei heavier than iron. A lot of energy released gravitationally during a SN, together with the presence of many neutrons and iron-peak elements, helps to synthesize heavier elements via the r-process in outer regions: successive neutrons are captured before there is time for beta decays, until the next neutron wouldn t be bound at all. Then the process hangs up until one of the nuclear neutron decays, when the process will continue.

89 rapid r-process a nucleosynthesis process, that occurs in core-collapse supernovae and is responsible for the creation of approximately half of the neutron-rich atomic nuclei heavier than iron. A lot of energy released gravitationally during a SN, together with the presence of many neutrons and iron-peak elements, helps to synthesize heavier elements via the r-process in outer regions: successive neutrons are captured before there is time for beta decays, until the next neutron wouldn t be bound at all. Then the process hangs up until one of the nuclear neutron decays, when the process will continue. The immediate products of this process are highly unstable nuclides and will decay, at leisure after the ejection, to the most neutron-rich stable isotopes of heavy elements like 176 Yb (ytterbium).

90 rapid r-process a nucleosynthesis process, that occurs in core-collapse supernovae and is responsible for the creation of approximately half of the neutron-rich atomic nuclei heavier than iron. A lot of energy released gravitationally during a SN, together with the presence of many neutrons and iron-peak elements, helps to synthesize heavier elements via the r-process in outer regions: successive neutrons are captured before there is time for beta decays, until the next neutron wouldn t be bound at all. Then the process hangs up until one of the nuclear neutron decays, when the process will continue. The immediate products of this process are highly unstable nuclides and will decay, at leisure after the ejection, to the most neutron-rich stable isotopes of heavy elements like 176 Yb (ytterbium). p-process proton capture of type (p,γ) in the high-energy proton field, is proposed to yield proton-rich isotopes of the elements from selenium to mercury. But their real origin is still not completely understood.

91 Periodic table and Nuclear Synthesis Color coded elements according to synthesis processes: in the first three minutes of the big bang (gray); fusion reactions in stars (burgundy); slow neutron capture, called the s-process, occurs in massive stars to form elements beyond iron (brown); rapid neutron-capture reaction called the r-process in explosive environments, such as in a supernova (turquoise).

92 Review Key concepts: Nuclear binding energy, nuclear reaction yield, binding fraction, compound nucleus, Gamow Peak, resonant and non-resonant reactions, pp-chain, CNO and triple-α processes, photo-disintegration, r-, s-, and p-processes, etc. 1. What is thermonuclear fusion? 2. What may you infer from the binding fraction vs. atomic mass number plot of nuclei (e.g., Fig. 6.1 in the textbook) qualitatively about both the nuclear force and the role of the electrostatic force between charged nucleons? 3. What is the Coulomb barrier of a nuclear reaction? What is the quantum tunneling? Why is it important in determining the rate of a reaction in a star? 4. What is the condition required for a resonant reaction? 5. How different are the dependence of resonant and non-resonant reaction rates on temperature (especially in the exponents)? 6. Why is the pace of the p-p chain or even the lifetime of the main sequence of a star very much determined by the rate of the p-p reaction? What force is involved in the reaction? 7. How is essentially all 14 N observed in nature produced in stars? 8. Under which condition, the nuclear Saha equation can be used? 9. What are α elements? Give a few examples. Which nuclear synthesis processes are mostly responsible for their production? 10. How are elements heavier than 56 Ni produced?